Question

The transverse displacement of a string fixed at both ends is given by
$y=0.06sin\left(\frac{2\pi x}{3}\right)cos\left(120\pi t\right)$
where x and y are in metres and t is in seconds. The length of the string is 1.5 m and its mass is $3.0\times {10}^{-2}kg$. What is the tension in the string?

• A. 648 N
• B. 724 N
• C. 832 N
• D. 980 N

Answer 1

The correct option is A 648 N

Let $\lambda$ be the wavelength, $f$ the frequency and $v$ the speed of the wave. Then $\frac{2\pi }{\lambda }=$ Coefficient of x in the argument of the sine funciton
$\Rightarrow \frac{2\pi }{\lambda }=\frac{2\pi }{3}$
$\Rightarrow \lambda =3m$
Also, $\omega =2\pi f$ = Coefficient of t in the argument of the cos funciton
$\Rightarrow 2\pi f=120\pi$
$\Rightarrow f=60Hz$
Hence, velocity of the wave, $v=f\lambda =60\times 3=180m{s}^{-1}$
Mass per unit length, $\mu =\frac{3.0\times {10}^{-2}}{1.5}=2.0\times {10}^{-2}kg{m}^{-1}$
We know that, $v=\sqrt{\frac{T}{\mu }}$, where T is tension in the string.
$\therefore T=\mu v^2=2.0\times {10}^{-2}\times (180{)}^2=648N$
Hence, the correct choice is (a).